How to use logarithm to solve for x?

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Can someone please explain to me how to do from question e onwards? Thanks!

2 Answers
Jul 13, 2017

Each of these log questions are written in the form a^(bx)=cabx=c, as log_a(b)^c-=clog_a(b)loga(b)ccloga(b) (the power rule), then log(a)^(bx)=log(c)-=bxlog(a)=log(c)log(a)bx=log(c)bxlog(a)=log(c), where x=log(c)/(blog(a))x=log(c)blog(a)," (change of base rule).

For example, let's take 2^(5x)=10025x=100. We can take the log of both sides to get log(2)^(5x)=log(100)log(2)5x=log(100), using the power law we get: 5xlog(2)=log(100)5xlog(2)=log(100), so x=log(100)/(5log(2))=1.32877124~~1.33x=log(100)5log(2)=1.328771241.33

Jul 14, 2017

(e)color(white)(x) 1+log_(10)11(e)x1+log1011

Explanation:

(e)(e)

"using the "color(blue)"laws of logarithms"using the laws of logarithms

•color(white)(x)logx^nhArrnlogxxlogxnnlogx

•color(white)(x)log_b b=1xlogbb=1

"given "10^x=110given 10x=110

"take " log_(10)" of both sides"take log10 of both sides

rArrlog_(10)10^x=log_(10)110log1010x=log10110

rArrxcancel(log_(10)10)^1=log_(10)(11xx10)

rArrx=log_(10)10+log_(10)11

rArrx=1+log_(10)11